From a61b998270e8f955c603de96775e072a555f735d Mon Sep 17 00:00:00 2001
From: Gregor Kleen <gkleen@yggdrasil.li>
Date: Tue, 19 Apr 2016 15:21:02 +0200
Subject: AlgoDat 01

---
 ss2016/algodat/01/H1-2.md | 10 ++++++++++
 1 file changed, 10 insertions(+)
 create mode 100644 ss2016/algodat/01/H1-2.md

(limited to 'ss2016/algodat/01/H1-2.md')

diff --git a/ss2016/algodat/01/H1-2.md b/ss2016/algodat/01/H1-2.md
new file mode 100644
index 0000000..7a50a2a
--- /dev/null
+++ b/ss2016/algodat/01/H1-2.md
@@ -0,0 +1,10 @@
+\begin{align*}
+  L &= \lim_{n \to \infty} \frac{n \cdot 2^n}{n^n} \\
+    &= \lim_{n \to \infty} \frac{2^{\log_2(n) + n}}{2^{\log_2(n) \cdot n}} \\
+    &= \lim_{n \to \infty} 2^{n + \log_2(n) - \log_2(n) \cdot n} \\
+  n + \log(n) - \log(n) \cdot n &\leq 2n - \log(n) \cdot n \\
+    &= (2 - \log(n)) \cdot n \xrightarrow{n \to \infty} - \infty \\
+  L &= 0
+\end{align*}
+
+$n \cdot 2^n$ und $n^n$ sind also sogar in der selben Komplexitätsklasse.
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